The health of a lion population in a safari park is monitored by measurements. a sample...
The health of a lion population in a safari park is monitored by measurements. a sample of 86 lions has a mean weight of 169.8 lb. Assume that the population standard deviation is known to be 46.5 lb and find a 93% confidence interval estimate of the mean of the population and find the margin of error.
Homework Answers
Solution :
Given that,
Point estimate = sample mean = 
= 169.8
Population standard deviation = 
= 46.5
Sample size = n = 86
At 93% confidence level
= 1 - 93%
= 1 - 0.93 =0.07
/2
= 0.035
Z/2
= Z0.035 = 1.812
Margin of error = E = Z/2
* ( 
/
n)
= 1.812 * ( 46.5 / 86
)
= 9.09
At 93% confidence interval estimate of the population mean is,
±
E
169.8 ± 9.09
(160.71, 178.89)
Margin of error = E = Z/2
* (
/n)
E = 1.812 * ( 46.5 / 86
)
E = 9.09
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